Log help again plzzzzzzzz

Express the following in the form lnx = ax + b and find a and b

1) xe^-x = 3.46 ans a=1 and b=0.9

2) (xe^x)^2 = 30e^-x ans a=-1.5 and b =1.7

thank u allot i really appreciate it coz i seriously cant solve this

ln (xe^-x)=ln3.46
ln x + lne^-x=ln3.46
lnx-xlne=ln3.46
lnx-x=ln3.46
lnx=x+ln3.46 =x+1.24 (which means: question probably waswith ln2.46=0.9)


ln(xe^x)^2= ln30e^-x
lnx^2+lne^2x=ln30e^-x
2lnx+2x=ln30+lne^-x
2lnx+2x=ln30 -x lne
2lnx=-3x+ln30
lnx=-3/2x+1.7

key to solution is: take ln from both sides of the equation. then calculate.

While #2 seems right, b in #1 ought to be ln 3.46 or 1.24127

Here is my work:
1) xe^-x = 3.46 ans a=1 and b=0.9 Take natural log each side.
(ln x) -x = ln 3.46; solving gives ln x = x + ln 3.46

2) (xe^x)^2 = 30e^-x Take natural log each side.
2lnx +2x = ln30 -x solving gives

lnx = [ ln30 -x -2x ] /2 = -(3/2)x + (ln30)/2 where your answers are OK.

no clue